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135 lbs to kg (Unit Conversion by Dimensional Analysis)

How many kilograms is 135 lbs (interpreting lb as pound-mass), and how is the conversion shown using dimensional analysis and significant figures?

Subject: Physics Classical Mechanics Chapter: Measurements Topic: Dimensional Analysis Answer included
135 lbs to kg lb to kg pounds to kilograms dimensional analysis unit cancellation conversion factor 0.45359237 significant figures mass vs weight
Accepted answer Answer included

Goal

Convert 135 lbs to kilograms using dimensional analysis (unit cancellation). In classical mechanics, “lb” can mean pounds-force (force) or pound-mass (mass); in a standard lbs to kg conversion, lb is interpreted as pound-mass.

Step 1: Use the exact conversion factor

The pound-mass is defined in terms of the kilogram by: \[ 1\ \text{lb} = 0.45359237\ \text{kg} \]

Step 2: Set up dimensional analysis (units cancel)

Multiply by a ratio equal to 1, arranged to cancel lb: \[ 135\ \text{lb} \times \frac{0.45359237\ \text{kg}}{1\ \text{lb}} \]

The lb unit cancels, leaving kilograms.

Step 3: Perform the calculation

\[ 135 \times 0.45359237 = 61.23496995 \] \[ 135\ \text{lb} = 61.23496995\ \text{kg} \]

Step 4: Round using significant figures

The value 135 has three significant figures. The conversion factor is exact, so the result is reported to three significant figures: \[ 61.23496995\ \text{kg} \approx 61.2\ \text{kg} \]

Final result: 135 lbs to kg gives \(61.2\ \text{kg}\) (3 significant figures).

Quick check (sanity check)

Since \(1\ \text{kg} \approx 2.20462\ \text{lb}\), converting back should return about 135 lb: \[ 61.23496995\ \text{kg} \times 2.20462\ \frac{\text{lb}}{\text{kg}} \approx 135\ \text{lb} \]

Visualization: one-step conversion with unit cancellation

135 lb (given) Multiply by 0.45359237 kg / 1 lb lb cancels → kg remains 61.2 kg (3 s.f.) Dimensional analysis
The conversion uses a unit ratio so that lb cancels, leaving the SI mass unit kg.

Optional mechanics note: weight as a force

If a mass \(m = 61.23496995\ \text{kg}\) is needed to estimate weight near Earth using \(W = m g\) and \(g = 9.80665\ \text{m/s}^2\), then: \[ W = 61.23496995 \times 9.80665 \approx 600.5095953\ \text{N} \]

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